C H A P. VI

OF THE RESOLUTION OF EQUATIONS, ALL WHOSE ROOTS ARE COMMENSURATE.

§ 53. I

T was demonftrated in Chap. 2. that the last term of any equation is the product of its roots: from which it follows, that the roots of an equation, when commenfurable quantities, will be found among the divifors of the laft term. And hence we have for the refolution of equations this

RULE.

Bring all the terms to one fide of the equation, find all the divifors of the last term, and fubftitute them fucceffively for the unknown quantity in the equation. So fhall that divifor which, fubftituted in this manner, gives the refulto, be the root of the propofed equa

tion.

For example, fuppofe this equation is to be refolved,

where the laft term is 2ab, whofe fimple lite

ral

ral divifors are a, b, 2a, 2b, each of which may, be taken either pofitively or negatively: but as here we find there are variations of figns in the equation, we need only take them pofitively. Suppose xa the firft of the divifors, and fubftituting a for x, the equation becomes

a'b}or,3a2—3a2+3a2b—3a2b=0.

a3-3a3 +2a3—2a'b —a2b+3a2b

So that, the whole vanishing, it follows that a is one of the roots of the equation.

After the fame manner, if you fubftitute b in place of x, the equation is

b3 — 3ab2 + 2a2b - 2a2b

24% } = 0,

which vanishing fhews b to be another root of the equation.

Again, if you fubftitute 2 a for x, you will find all the terms destroy one another fo as to make the fumo. For it will then be,

Whence we find that 2a is the third root of the equation. Which, after the first two (+ a, + b) had been found, might have been collected from this, that the laft term being the product of the three roots, + a, + b being known,

the third must neceffarily be equal to the laft term divided by the product ab, that is, =

2a2b

ab

= 2a.

Let the roots of the cubic equation

2x

33*+90=o be required. And firft the divifors of 90 are found to be 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90. If you substitute 1 for x, you will find ×3 I 33% +90=56; fo that I is not a root of the equation. If you fubftitute 2 for x, the refult will be 24: but putting x = 3, you have x3−2x2−33x+90=27—18—99+90=117-117=0. So that 3 is one of the roots of the propofed equation. The other affirmative root is 5; and after you find it, as it is manifeft from the equation, that the other root is negative, you are not to try any more divifors taken positively, but to fubftitute them, negatively taken, for x: and thus you find that 6 is the third root. For putting = - 6, you have

x3-2x2-33x+90=-216-72+ 198+90=0.

This last root might have been found by dividing the laft term 90, having its fign changed, by 15, the product of the two roots already found.

$55. When one of the roots of an equation is found, in order to find the rest with lefs trouble, divide the propofed equation by the fimple equation which you are to deduce from the root

already

already found, and the quotient shall give an equation of a degree lower than the proposed; whofe roots will give the remaining roots required.

As for example, the root + 3, first found, gave 3 or x-3=0, whence dividing thus, x

-X

X3

33x+90(x2 + x — 30

The quotient fhall give a quadratic equation x2 + x―30= o, which must be the product of the other two fimple equations from which the cubic is generated, and whofe roots therefore must be two of the roots of that cubic.

Now the roots of that quadratic equation are eafily found by Chap. 13. Part I. to be + 5 and -6. For,

3

$56. After the fame manner, if the biquadratic + 2x3 25x2 + 26x + 1200 is to. be refolved; by fubftituting the divifors of 120 for x, you will find that + 3, one of those divisors, is one of the roots; the fubftitution of for x giving 81 54 → 225 + 78 + 120 = 279-279-0. And therefore dividing the propofed equation by x3, you must enquire for the roots of the cubic x3 + x2 22x 40 = 0, and finding that +5, one of the divifors of 40, is one of the roots, you divide that cubic by x-5, and the quotient gives the quadratic x2 + 6x + 8 = o, whose two roots are -2, -4. So that the four roots of the biquadratic are + 3, + 5,

2

2,

- 4.

§ 57. This Rule fuppofes that you can find all the divifors of the laft term; which you may always do thus.

bave

If it is a fimple quantity, divide it by its leaft divifor that exceeds unit, and the quotient again by its leaft divifor, proceeding thus till you a quotient that is not divifible by any number greater than unit." This quotient, with these divifors, are the first or fimple divifors of the quantity. And the products of the multiplication of any 2, 3, 4, &c. of them are the compound divifors.

As, to find the divifors of 60; firft I divide by 2, and the quotient 30 again by 2, then the